Quantum mechanics - spin operator eigenvalue probability?

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  • #1
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Hi,
I have this problem on a past exam paper I am having some trouble with:

"in the conventional basis of the eigenstates of the Sz operator, the spin state of a spin-1/2 particle is described by the vector:

[tex] u = \left( \stackrel{cos a}{e^i^b sina} \right)[/tex] where a and B are constants.

find the probability that a measurement of the y-component of the spin of the particle will yield the result [tex] 0.5\hbar [/tex] ."

For the life of me I cannot work out how to write out matrices legibly on this thing, so I will summarize what is bothering me. I am given pauli matrices [tex] \sigma_x_,_y_,_z [/tex]that I cannot write out properly, and the spin operator is given by [tex] S_i = i\hbar\sigma_i [/tex].

In the question I am given the vector u, which is apparently expressed in the basis of Sz eigenstates.

Am I justified in putting this vector u into an eigenvalue equation [tex] S_{y} u = a_{y} u [/tex] ,

where ay is my eigenvalue, when the vector I would be operating on is made from a basis of eigenstates of another operator (Sz)?

I tried this and got two equations for ay, neither of which gives [tex] a_y = 0.5\hbar [/tex].

does this mean I can conclude that there is zero probability of finding the y-component of the spin being equal to [tex] 0.5\hbar [/tex] ?

or do I somehow have to wangle it so that I get another vector (not u) that is in the Sy eigenstate basis?

thanks.



PS. sorry, this crazy thing will not let me change something 5 lines up where I should have said
"Am I justified in putting this vector u into an eigenvalue equation S_y u = 0.5\hbar u"
 
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  • #2
gabbagabbahey
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"Am I justified in putting this vector u into an eigenvalue equation S_y u = 0.5\hbar u"
You tell me. If [itex]u[/itex] is an eigenvector of [itex]S_z[/itex], is it also an eigenvector of S_y[/itex]?
 
  • #3
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You tell me. If [itex]u[/itex] is an eigenvector of [itex]S_z[/itex], is it also an eigenvector of S_y[/itex]?
are you actually asking me, do these two operators commute? and if their commutator [Sy, Sz] = 0, then yes they do?

is that right, if the eigenvector of one operator is equal to the eigenvector of another, then the operators commute?
 
  • #4
gabbagabbahey
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if the eigenvector of one operator is equal to the eigenvector of another, then the operators commute?
Yes. So, do [itex]S_z[/itex] and [itex]S_y[/itex] commute?
 
  • #5
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Yes. So, do [itex]S_z[/itex] and [itex]S_y[/itex] commute?
not by my calculations, although I just woke up 5 minutes ago. I get [Sy,Sz] is equal to
(hbar^2)/4 multiplied by a 2x2 matrix containing zeros on the leading diagonal and 2i elsewhere.
 
  • #6
gabbagabbahey
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Right, they don't commute; [itex][S_y,S_z]=i\hbar S_x[/itex]. So, [itex]u[/itex] is clearly not an eigenstate of [itex]S_y[/itex].

What are the eigenstates of [itex]S_y[/itex]?
 
  • #7
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Right, they don't commute; [itex][S_y,S_z]=i\hbar S_x[/itex]. So, [itex]u[/itex] is clearly not an eigenstate of [itex]S_y[/itex].

What are the eigenstates of [itex]S_y[/itex]?
i got that one of them was the column vector, say, x:
x_1

ix_1

and the other vector y:
y_1

-iy_1

but I was not convinced by this because, if the upper component is say, x1, and I choose x1 = 1, then if I go to normalize it i do
1/sqrt[(1^2) + (i^2)] = 1/(1-1) = 0, so how can this be right if Im getting a normalization constant of zero?
 
  • #8
vela
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You forgot to take the conjugate. You should be calculating [itex]\langle x|x \rangle = x^\dagger x[/itex], not [itex]x^T x[/itex].
 

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